Graph the conic section x28y1 include foci vertices symptote

Graph the conic section: x^2=-8(y-1) (include foci, vertices, symptotes when required

Solution

This conic section is a vertical parabola. The vertex form of a vertical parabola is
4p(y-k) = (x-h)², where the vertex is (h,k) and the distance from the vertex to the focus is p.

We have
-8(y-1)=x²,
so the vertex is on the y-axis at (0,1) and, since p=-2, the focus is 2 units below the vertex, at (0,-1), so the parabola opens downwards.

To graph the parabola, which is symmetric about the y-axis, I would write it in standard form,
y=(-1/8)x²+1,
and pick a few positive x-values and calculate y.